470 replies to this topic

[Finix]

I heard a similar one on khan academy

[Morningstar]

Is this actually still being discussed? It was started at the end of 2007!

[zorrotornado]

easy one come on

Spoiler for please use spoilers

19 and 50% for the first

the night before they agree that the first will count lets say the red hats, if the number is odd he will say red, if the number is even he will say black
lets say he says black and dies, the 2nd will cound how many hats he has in front of him. since the 1 said black it means that there is 0,2,4,6,8,10...... red hats, soo the 2nd counts the number of hats in front of him, if he finds an odd number it means that he has a black one if its an even number it means he has a red one . and it continues until the last .

Edited by plainglazed, 23 November 2011 - 11:42 PM.

[bendigi]

I do have another solution skipping the math. Although, the odd/even solution is quite nice.

I would love to get some feedback on my solution:

The first guy that is being asked, simply tells the color of the man in front of him, if he is lucky he lives, if not, oh well he did it for the team.
Now, the next guy has already heard his color (black or red) and he is guaranteed freedom, but to help the next guy, here is my idea
If his color matches the one in front, he says the color without hesitation, if it's different, he pauses and thinks and then says his color.
The next guy listening, will consider the pause as a negation of the prevoius color and continue this process ...

Good one ?

[Swinslow]

Your reasoning is correct. It was stated that a correct strategy would guarantee the safety of 19 of the 20, with the 20th (first to guess) having a 50% chance.

With no other information present, I don't see how this is solvable. What strategy would allow you to say the word "red" or the word "black" and tell nineteen people which of two colors each was wearing? I believe it is impossible; you cannot convey that much information in a simple binary choice.

I can see a method to save at least 13 of the 20 people:

Spoiler for solution

The first person looks at the two hats in front of him. If the hats are the same color, he says "red". If they are different colors, he says "black". This guarantees the safety of the two people in front of him, since the next guy in line will be able to figure out his own color. Starting with the first, every third person uses this code. There will be two people left at the end; the second-to-last guesses the color of the last person's hat, so the last person will also be assured of survival. If done correctly, only seven of the twenty will be in danger of death. A mathematically average run would see 3 or 4 executions among the 20, a 15-20% mortality rate -- not bad odds for 20 condemned prisoners. At worst, 13 survive, for a 65% minimum survival rate. And there is one chance in 128 that all will survive.

I see but knowing prisoners.....

Spoiler for

they will find a way of comminicating the colours through coughing, sneezing etc. this will mean that only one prisoner will have a chance of death resulting if he cooperates the 19 others an almost guarrentted escape

[Tyler Durden]

Sorry. I must have missed something in the instructions. This solution only works when there are 10 black hats and 10 red hats. No where in the instructions does it say this. The solution fails if there are different numbers of black and red hats. There could be 19 black hats and 1 red hat.

[Chew Jing Kai]

All 20 prisoners will be freed

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20

Prisoner 20 will start first. Since he can see everyone in front of him, he knows and will shout the color of his hat (if he sees 9 black 10 red that means he is wearing black, and vice versa)
Prisoner 19 will also be able to shout the color of his hat (if he sees 9 black 9 red and prisoner 20 shouted black, that means he is wearing red)
Prisoners 18 to 2 repeat the same steps.
Prisoner 1 having heard all the color shouted before his turn will also know his color (if 9 shouted black and 10 shouted red that means he is black)

[Ranbir Singh]

Here's my solution, not sure if it's allowed or not...

Spoiler for Here's my solution.. probably something wrong with it...

You can save the 19 ahead of you by stretching out how long you say the color. The guy in the very back (20th in line) would say the color of the individual (19th) in front of him in 1 second. (Red.) He's got the 50-50 chance. The next person, if 18 is red as well, he says it in one second. If not, he counts off the number of people until he sees another red and that's how long it takes him to say "Red." So if the person 3 (number 16) ahead of him has a red hat, he says "Reeeeeed" over 3 seconds. This way, everyone knows counts the number of seconds for the word and knows that if it skips them, they are the opposite color. If you get close to the end and there are 3 people left with red hats and yours is black, you extend it to 4 seconds for saying red, telling the three of them they are all black.

How's this one?

one wrong timing and all are gone !!

[Brainiac101]

1.If the last person is right you say the opposite color if he is wrong you say the same color
2.You all say the same color and hope your right

[vivekkumarjha]

well, nice one
but the slightest mistake and all would be executed

Edited by vivekkumarjha, 13 June 2012 - 05:39 PM.